Chapter 23 Wings 607 Chapter 24 Fuselage frames and wing ribs 638
21.3 Beams having variable stringer areas
21.3 Beams having variable stringer areas
In many aircraft, structural beams, such as wings, have stringers whose cross-sectional areas vary in the spanwise direction. The effects of this variation on the determina- tion of shear flow distribution cannot therefore be found by the methods described in Section 20.3 which assume constant boom areas. In fact, as we noted in Section 20.3, if the stringer stress is made constant by varying the area of cross-section there is no change in shear flow as the stringer/boom is crossed.
The calculation of shear flow distributions in beams having variable stringer areas is based on the alternative method for the calculation of shear flow distributions described in Section 20.3 and illustrated in the alternative solution of Example 20.3. The stringer loads Pz,1and Pz,2are calculated at two sections z1 and z2of the beam a convenient distance apart. We assume that the stringer load varies linearly along its length so that the change in stringer load per unit length of beam is given by
P= Pz,1−Pz,2 z1−z2
The shear flow distribution follows as previously described.
Example 21.3
Solve Example 21.2 by considering the differences in boom load at sections of the beam either side of the specified section.
In this example the stringer areas do not vary along the length of the beam but the method of solution is identical.
We are required to find the shear flow distribution at a section 2 m from the built-in end of the beam. We therefore calculate the boom loads at sections, say 0.1 m either side of this section. Thus, at a distance 2.1 m from the built-in end
Mx = −100×1.9= −190 kN m
The dimensions of this section are easily found by proportion and are width=1.18 m, depth=0.59 m. Thus the second moment of area is
Ixx=4×900×2952+2×1200×2952=5.22×108mm4 and
σz,r= −190×106
5.22×108 yr= −0.364yr Hence
P1 =P3= −P4= −P6= −0.364×295×900= −96 642 N and
P2= −P5= −0.364×295×1200= −128 856 N
At a section 1.9 m from the built-in end
Mx = −100×2.1= −210 kN m
and the section dimensions are width = 1.22 m, depth = 0.61 m so that Ixx=4×900×3052+2×1200×3052=5.58×108mm4 and
σz,r= −210×106
5.58×108 yr= −0.376yr Hence
P1 =P3= −P4= −P6= −0.376×305×900= −103 212 N and
P2= −P5 = −0.376×305×1200= −137 616 N
Thus, there is an increase in compressive load of 103 212−96 642=6570 N in booms 1 and 3 and an increase in tensile load of 6570 N in booms 4 and 6 between the two sec- tions. Also, the compressive load in boom 2 increases by 137 616−128 856=8760 N while the tensile load in boom 5 increases by 8760 N. Therefore, the change in boom load per unit length is given by
P1=P3= −P4= −P6= 6570
200 =32.85 N and
P2= −P5= 8760
200 =43.8 N
The situation is illustrated in Fig. 21.9. Suppose now that the shear flows in the panels 12, 23, 34, etc. are q12, q23, q34, etc. and consider the equilibrium of boom 2, as shown in Fig. 21.10, with adjacent portions of the panels 12 and 23. Thus
q23+43.8−q12=0 or
q23=q12−43.8 Similarly
q34=q23−32.85=q12−76.65 q45=q34+32.85=q12−43.8 q56=q45+43.8=q12
q61=q45+32.85=q12+32.85
21.3 Beams having variable stringer areas 595
Fig. 21.9Change in boom loads/unit length of beam.
Fig. 21.10Equilibrium of boom.
The moment resultant of the internal shear flows, together with the moments of the components Py,r of the boom loads about any point in the cross-section, is equivalent to the moment of the externally applied load about the same point. We note from Example 21.2 that for moments about the centre of symmetry
6 r=1
Px,rηr=0 6 r=1
Py,rξr=0
Therefore, taking moments about the centre of symmetry
100×103×600=2q12×600×300+2(q12−43.8)600×300
+(q12−76.65)600×600+(q12+32.85)600×600 from which
q12=62.5 N/mm
whence
q23=19.7 N/mm q34= −13.2 N/mm q45=19.7 N/mm, q56=63.5 N/mm q61=96.4 N/mm
so that the solution is almost identical to the longer exact solution of Example 21.2.
The shear flows q12, q23, etc. induce complementary shear flows q12, q23, etc. in the panels in the longitudinal direction of the beam; these are, in fact, the average shear flows between the two sections considered. For a complete beam analysis the above procedure is applied to a series of sections along the span. The distance between adjacent sections may be taken to be any convenient value; for actual wings distances of the order of 350–700 mm are usually chosen. However, for very small values small percentage errors in Pz,1and Pz,2result in large percentage errors inP. On the other hand, if the distance is too large the average shear flow between two adjacent sections may not be quite equal to the shear flow midway between the sections.
Problems
P.21.1 A wing spar has the dimensions shown in Fig. P.21.1 and carries a uniformly distributed load of 15 kN/m along its complete length. Each flange has a cross-sectional area of 500 mm2with the top flange being horizontal. If the flanges are assumed to resist all direct loads while the spar web is effective only in shear, determine the flange loads and the shear flows in the web at sections 1 and 2 m from the free end.
Ans. 1 m from free end: PU=25 kN (tension), PL=25.1 kN (compression), q=41.7 N/mm.
2 m from free end: PU=75 kN (tension), PL=75.4 kN (compression), q=56.3 N/mm.
Fig. P.21.1
P.21.2 If the web in the wing spar of P.21.1 has a thickness of 2 mm and is fully effective in resisting direct stresses, calculate the maximum value of shear flow in the web at a section 1 m from the free end of the beam.
Ans. 46.8 N/mm.
Problems 597 P.21.3 Calculate the shear flow distribution and the stringer and flange loads in the beam shown in Fig. P.21.3 at a section 1.5 m from the built-in end. Assume that the skin and web panels are effective in resisting shear stress only; the beam tapers symmetrically in a vertical direction about its longitudinal axis.
Ans. q13=q42=36.9 N/mm, q35=q64=7.3N/mm, q21=96.2 N/mm, q65=22.3 N/mm.
P2= −P1=133.3 kN, P4=P6= −P3= −P5=66.7 kN.
Fig. P.21.3
22
Fuselages
Aircraft fuselages consist, as we saw in Chapter 12, of thin sheets of material stiffened by large numbers of longitudinal stringers together with transverse frames. Gener- ally, they carry bending moments, shear forces and torsional loads which induce axial stresses in the stringers and skin together with shear stresses in the skin; the resistance of the stringers to shear forces is generally ignored. Also, the distance between adjacent stringers is usually small so that the variation in shear flow in the connecting panel will be small. It is therefore reasonable to assume that the shear flow is constant between adjacent stringers so that the analysis simplifies to the analysis of an idealized section in which the stringers/booms carry all the direct stresses while the skin is effective only in shear. The direct stress carrying capacity of the skin may be allowed for by increasing the stringer/boom areas as described in Section 20.3. The analysis of fuse- lages therefore involves the calculation of direct stresses in the stringers and the shear stress distributions in the skin; the latter are also required in the analysis of transverse frames, as we shall see in Chapter 24.